Question

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streaming is gaining in popularity. A poll reported that 55% of 2341 American adults surveyed said they have watched digitally streamed TV programming on some type of device.

(a)

Calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time. (Round your answers to three decimal places.) , Interpret the resulting interval.

We are 99% confident that this interval does not contain the true population proportion.


We are 99% confident that this interval contains the true population proportion.


We are 99% confident that the true population proportion lies below this interval.We are 99% confident that the true population proportion lies above this interval.

Answer 1

Answer:

a) The 99% confidence interval would be given (0.523;0.577).

We are 99% confident that this interval contains the true population proportion.

b) $n=\frac{0.55(1-0.55)}{(\frac{0.03}{2.58})^2}=1830.51$  

And rounded up we have that n=1831

Step-by-step explanation:

Data given and notation  

n=2341 represent the random sample taken    

X represent the people that they have watched digitally streamed TV programming on some type of device

$\hat p=0.55$ estimated proportion of people that they have watched digitally streamed TV programming on some type of device  

$\alpha=0.01$ represent the significance level

Confidence =0.99 or 99%

z would represent the statistic for the confidence interval  

p= population proportion of people that they have watched digitally streamed TV programming on some type of device

The population proportion present the following distribution:

$p \sim N (p, \sqrt{\frac{p(1-p)}{n}}$

Part a) Confidence interval

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.55 - 2.58 \sqrt{\frac{0.55(1-0.55)}{2341}}=0.523$

$0.55 + 2.58 \sqrt{\frac{0.55(1-0.55)}{2341}}=0.577$

And the 99% confidence interval would be given (0.523;0.577).

We are 99% confident that this interval contains the true population proportion.

Part b) What sample size would be required for the width of a 99% CI to be at most 0.03 irrespective of the value of p??

The margin of error for the proportion interval is given by this formula:  

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$    (a)  

And on this case we have that $ME =\pm 0.03$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$   (b)  

And replacing into equation (b) the values from part a we got:

$n=\frac{0.55(1-0.55)}{(\frac{0.03}{2.58})^2}=1830.51$  

And rounded up we have that n=1831